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Lady_Fox [76]
2 years ago
8

If 4x-5y=6 and xy=8 then find the value of 16x^2+25y^2​

Mathematics
1 answer:
Anit [1.1K]2 years ago
6 0

Answer:

Hello,

356

Step-by-step explanation:

4x-5y=6\\xy=8\\\\(4x-5y)^2=16x^2+25y^2-40xy\\\\\\16x^2+25y^2=(4x-5y)^2+40xy=6^2+40*8=356\\

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What is the equation of the line that passes
Afina-wow [57]

Answer:

Step-by-step explanation:

First, find the slope of the line for the two points.

Slope is Rise/Run.

For the two points (-8,-4) and (3,-10):

Rise (-10 - (-4)) = -6

Run (3 - ( -8)) = 11

The slope is Rise/Run or -(6/11)

The equation becomes y = -(6/11)x + b

Find b by entering either of the 2 points given.  I'll use (3,-10):

y = -(6/11)x + b

-10 = -(6/11)(3) + b

b = -10 + (6/11)(3)

b = -10 +(18/11)

b = -(110/11) + (18/11)

b = -(92/11)

The equation becomes y = -(6/11)x - (92/11)

See attached graph.

4 0
2 years ago
(3 + 4i) + (5 − 2i) (2 points) −2 + 6i 2 − 2i 7 + 3i 8 + 2i
k0ka [10]

Answer:

8+2i

Step-by-step explanation:

Combine like terms

3+5=8

4i-2i=2i

6 0
3 years ago
Just to ensure I'm doing this method right, (for clarification)
EleoNora [17]
You are right i tried it and i got it right
4 0
3 years ago
Which is the better buy, 7 pounds for $8.47 or 9 pounds for $11.07?
wel
Let us take the case of 7 pounds for $8.47 first.
7 pounds of a product costs = 8.47 dollars
Then
1 pound of the same product will cost = (8.47/7) dollars
                                                             = 1.21 dollars
Now let us take the case of 9 pounds for $11.07
9 pounds of a product costs = 11.07 dollars
Then
1 pound of the same product will cost = (11.07/9) pounds
                                                             = 1.23 dollars
So from the above deductions we can see that 9 pounds for $11.07 is a better buy than 7 pounds for $8.47. I hope the procedure is clear enough for you to understand. In future you can use this method for solving similar problems.
7 0
2 years ago
Read 2 more answers
Guided Practice
marysya [2.9K]
It’s d because 12-15%=10.20
7 0
2 years ago
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